3 Answers
3

We can get a somewhat different point of view by looking at arclength, surface area, and other problems for curves given parametrically.

Let the curve be given by $x=x(t)$, $\:y=y(t)$. Then the surface area obtained when we rotate the chunk of the curve from $t=a$ to $t=b$ around the $x$-axis is
$$\int_{t=a}^b 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$
For rotation about the $y$-axis, we have a very similar expression for surface area:
$$\int_{t=a}^b 2\pi x\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$

In our case, we have $y=\sqrt[3]{x}+2$. Let's choose a nice parametrization, something that will simplify the calculations. A sensible choice is $x=t^3$, in which case $y=t+2$. Since $x$ goes from $1$ to $8$, $t$ will go from $1$ to $2$.

Calculate. We have $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=1$.
For rotation about the $y$-axis, we get surface area
$$\int_1^2 2\pi t^3\sqrt{1+9t^4}\,dt.$$
This integral can be calculated by a simple substitution. But this was not your problem.

For rotation about the $x$-axis, we get surface area
$$\int_1^2 2\pi (t+2)\sqrt{1+9t^4}\,dt.$$
This is a nightmarish integral. The hardest part, actually the marginally easier $\int\sqrt{1+x^4}\,dx$, has been discussed on this site. It may not be worth looking up.

Whoever was writing out solutions for your book got lucky. If (s)he had not made a mistake, (s)he would have had an awful integral to evaluate. (It would start out looking even worse without the parametric approach.)

The formula you have provided assumes that the surface is being produced by rotating around the x-axis. In that case, f(x) is the radius of rotation. It looks like you might instead be creating the surface by rotating around the y-axis, in which case the radius of rotation would be x, as in the solution. You can think of finding the surface area as a bunch of little slanty sticks each carving out a thin strip of surface as they're carried around the axis of rotation. Each stick (represented by the stuff in the square root) traces out a strip of circumference 2$\pi$r, where r depends on the set-up of the rotation.